Question

If the function $f(x)$ is given by \[ f(x) = \begin{cases} \frac{\tan(a(x-1))}{\frac{x-1}{x}}, & tif0<x<1 
\frac{x^3-125}{x^2 - 25} , & \text{if } 1 \leq x \leq 4 
\frac{b^x - 1}{x}, & \text{if } x>4 \end{cases} \] is continuous in its domain, then find $6a + 9b^4$.

• A. 284
• B. 261
• C. 214
• D. 317 \bigskip

Hint:

When solving continuity problems, always check the limits from both sides at the points of discontinuity and equate them to ensure continuity.

Answer 1

The Correct Option is A

The given function $f(x)$ is: \[ f(x) = \begin{cases} \frac{\tan(a(x-1))}{\frac{x-1}{x}}, & \text{if } 0<x<1
\frac{x^3-125}{x^2 - 25} , & \text{if } 1 \leq x \leq 4
\frac{b^x - 1}{x}, & \text{if } x>4 \end{cases} \] To ensure continuity, we must check at $x = 1$ and $x = 4$. --- Step 1: Continuity at $x = 1$ For continuity at $x = 1$, we need: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$ # Left-hand limit ($x \to 1^-$): $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{\tan(a(x-1))}{\frac{x-1}{x}}$$ For small $x - 1$, using the approximation $\tan u \approx u$: $$\tan(a(x-1)) \approx a(x-1)$$ Thus, $$\frac{\tan(a(x-1))}{\frac{x-1}{x}} \approx \frac{a(x-1)}{\frac{x-1}{x}} = a \cdot x$$ Taking $x \to 1$: $$\lim_{x \to 1^-} f(x) = a(1) = a$$ # Right-hand limit and $f(1)$: $$f(1) = \frac{1^3 - 125}{1^2 - 25} = \frac{1 - 125}{1 - 25} = \frac{-124}{-24} = \frac{31}{6}$$ Equating limits: $$a = \frac{31}{6}$$ --- Step 2: Continuity at $x = 4$ For continuity at $x = 4$: $$\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4)$$ # Left-hand limit ($x \to 4^-$) and $f(4)$: $$f(4) = \frac{4^3 - 125}{4^2 - 25} = \frac{64 - 125}{16 - 25} = \frac{-61}{-9} = \frac{61}{9}$$ # Right-hand limit ($x \to 4^+$): $$\lim_{x \to 4^+} \frac{b^x - 1}{x} = \frac{b^4 - 1}{4}$$ Equating: $$\frac{b^4 - 1}{4} = \frac{61}{9}$$ Solving for $b$: $$b^4 - 1 = \frac{244}{9}$$ $$b^4 = \frac{253}{9}$$ $$b = \left(\frac{253}{9}\right)^{\frac{1}{4}}$$ --- Step 3: Compute $6a + 9b^4$ $$6a + 9b^4 = 6 \times \frac{31}{6} + 9 \times \frac{253}{9}$$ $$= 31 + 253 = 284$$ --- Final Answer: $$\boxed{284}$$ \bigskip